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Varvara68 [4.7K]
2 years ago
15

Chemical analysis shows that citric acid contains 37.51% C, 4.20% H, and 58.29% O. What is the empirical formula for citric acid

?​
Chemistry
1 answer:
Fiesta28 [93]2 years ago
7 0

The empirical formula for the citric acid is C₆H₈O₇

<h3>Data obtained from the question </h3>
  • Carbon (C) = 37.51%
  • Hydrogen (H) = 4.20%
  • Oxygen (O) = 58.29%
  • Empirical formula =?

Divide by their molar mass

C = 37.51 / 12 = 3.126

H = 4.2 / 1 = 4.2

O = 58.29 / 16 = 3.643

Divide by the smallest

C = 3.126 / 3.126 = 1

H = 4.2 / 3.126 = 1.34

O = 3.643 / 3.126 = 1.17

Multiply through by 6 to express in whole number

C = 1 × 6 = 6

H = 1.34 × 6 = 8

O = 1.17 × 6 = 7

Thus, the empirical formula for the citric acid is C₆H₈O₇

Learn more about empirical formula:

brainly.com/question/24818135

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Answer:

\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}

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<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

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There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

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Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}

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3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}

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<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}

Multiply by the number of moles we calculated.

0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}

The units of moles of sodium chloride cancel.

0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }

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The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

3.7 * 10^{22} \ molecules \ NaCl

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